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a diverging lens forms an image 1/3 rd the size of an object that is 24 cm away from the lens. determine the focal length of the lens.
Diverging lens : object height = h image height = h/3 object distance (u) = - 24 cm v = ? f = ? m = H/h =h/3/h = h/3h = 1/3 m = 1/3 m = -v/u 1/3 = -v/-24 v = 8 cm Now by lens formulae, 1/f = 1/v -1/u 1/f = 1/8 - (-1/24) 1/f = 1/8 + 1/24 = 3 + 1/24 = 4/24 = 1/6 f = 6 cm Hence the focal length of the lens will be 6 cm. . . Pleazzzzzzzz approveeeee
Diverging lens :
object height = h
image height = h/3
object distance (u) = - 24 cm
v = ?
f = ?
m = H/h =h/3/h = h/3h = 1/3
m = 1/3
m = -v/u
1/3 = -v/-24
v = 8 cm
Now by lens formulae,
1/f = 1/v -1/u
1/f = 1/8 - (-1/24)
1/f = 1/8 + 1/24
= 3 + 1/24
= 4/24
= 1/6
f = 6 cm
Hence the focal length of the lens will be 6 cm.
.
Pleazzzzzzzz
approveeeee
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